Problem 87 - Calculus and its applications

Question

The median weight of a boy whose age is between 0 and 36 months is approximated by $$ w(t)=8.15+1.82 t-0.0596 t^{2}+0.000758 t^{3} $$ where \(t\) is in months and \(w\) is in pounds. Use this approximation to find the following: a) The rate of change of weight with respect to time b) The weight of a boy at age 10 months c) The rate of change of a boy's weight with respect to time at age 10 months

Step-by-Step Solution

Answer

a) Rate of change: \(1.82 - 0.1192t + 0.002274t^2\), b) Weight at 10 months: 21.148 pounds, c) Rate at 10 months: 0.8554 pounds/month.

Step 1: Differentiate the Weight Function

To find the rate of change of weight with respect to time, we need to differentiate the function \( w(t) = 8.15 + 1.82t - 0.0596t^2 + 0.000758t^3 \).Differentiating, we have: \[ w'(t) = \frac{d}{dt} (8.15 + 1.82t - 0.0596t^2 + 0.000758t^3). \]Applying the power rule, the derivative is:\[ w'(t) = 0 + 1.82 - 2(0.0596)t + 3(0.000758)t^2 = 1.82 - 0.1192t + 0.002274t^2. \]

Step 2: Calculate Weight at 10 Months

Substitute \( t = 10 \) into the original function to find the weight:\[ w(10) = 8.15 + 1.82(10) - 0.0596(10)^2 + 0.000758(10)^3. \]Calculate each term:- \( 1.82(10) = 18.2 \)- \( 0.0596(100) = 5.96 \)- \( 0.000758(1000) = 0.758 \)Substitute and simplify:\[ w(10) = 8.15 + 18.2 - 5.96 + 0.758 = 21.148 \text{ pounds}. \]

Step 3: Calculate Rate of Change at 10 Months

Substitute \( t = 10 \) into the derivative to find the rate of change at 10 months:\[ w'(10) = 1.82 - 0.1192(10) + 0.002274(10)^2. \]Calculate each term:- \( 0.1192(10) = 1.192 \)- \( 0.002274(100) = 0.2274 \)Substitute and simplify:\[ w'(10) = 1.82 - 1.192 + 0.2274 = 0.8554 \text{ pounds per month}. \]

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus where we calculate the rate at which a function changes at any given point. This notion of rate of change is symbolized by the derivative. In the given problem, we need to determine how a boy's weight shifts with time, expressed by the function: \[ w(t) = 8.15 + 1.82t - 0.0596t^2 + 0.000758t^3. \]To find the derivative, or simply the rate at which weight changes with time \( t \), we differentiate the entire equation with respect to \( t \). This process involves taking each term of the polynomial and applying the power rule, a vital technique in differentiation: - For a constant like \( 8.15 \), the derivative is zero.- For \( 1.82t \), the derivative is simply the coefficient \( 1.82 \), as the derivative of \( t \) is 1.- For the term \( -0.0596t^2 \), applying the power rule gives us \(-2*0.0596t = -0.1192t \).- For \( 0.000758t^3 \), differentiation results in \(3*0.000758t^2 = 0.002274t^2 \). The derivative then becomes: \[ w'(t) = 1.82 - 0.1192t + 0.002274t^2 \] This new function \( w'(t) \) tells us how fast the weight of the boy is changing at any given month \( t \).

Mathematical Modeling

Mathematical modeling involves creating mathematical representations of real-world phenomena. Here, the problem uses a polynomial function to approximate the median weight of a boy from birth to 36 months. This representation is a typical application of mathematical models in successfully predicting or estimating outcomes by using mathematical formulas.In this particular scenario, the function \( w(t) = 8.15 + 1.82t - 0.0596t^2 + 0.000758t^3 \) acts as a predictive tool to estimate a boy's weight over a specific time span using input values for \( t \). Such models are essential tools in various fields like:

• Biology, to model growth processes.
• Finance, to predict economic trends.
• Engineering, for designing structures and systems.

Using a mathematical model provides essential insights and aids in decision-making processes. This equation specifically allows us to analyze weight development by modeling the weight gain pattern influenced by age.

Applied Mathematics

Applied mathematics is about utilizing mathematical methods to solve practical problems from various real-world domains. In the exercise, the aim is to apply mathematical calculations to get meaningful insights about the weight change of a young boy over time.The weight function \( w(t) \) not only provides a formulaic representation of growth but also allows for:- Calculating the exact weight at a particular age, such as at 10 months.- Estimating the rate of weight change over specific time intervals, as done when substituting the value of \( t = 10 \) into the derivative \( w'(t) \).Having a calculated rate of \( 0.8554 \) pounds per month at 10 months gives parents, doctors, or researchers invaluable information about the pace of growth at that point. Applied mathematics takes abstract mathematical computations and converts them into practical, usable information to help make real-world decisions based on quantifiable data.